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Galileo's rule of falling bodies

  1. Sep 14, 2011 #1
    Simple question, if the force of gravity is proportional to mass and distance, as in Newton's and Einstein's theories of gravitation, then why do people say that objects with two different masses accelerate at the same rate independent of mass on earth? Can Galileo's rule just be a practical approximation that doesn't pertain to reality?
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  3. Sep 14, 2011 #2


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    Relative to some inertial frame of reference, the rate of acceleration of an object in the direction of the earth is independent of the mass of that object, but the rate of acceleration of earth in the direction of the object will be affected by the mass of that object.

    In most cases, the earth is so massive compared to the object that it's rate of acceleration is tiny and can be ignored, and an earth bound observer will see virtually the same rate of acceleration of an object as an observer from an inertial (zero acceleration) frame of reference.
  4. Sep 14, 2011 #3

    Ken G

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    I'm not sure the response of the Earth is what chris2112 is asking, it sounds like a more basic issue about when do two different forces produce the same motion. He might be thinking that two different forces must yield two different accelerations, but remember, the acceleration obeys a = F/m, so two different F can give the same a if the m is different in the same way that the F is. That's exactly what happens in Newton's gravity-- an object with twice the mass gets twice the force, but an object with twice the mass has twice the inertia so needs twice the force to get the same acceleration. (Also a few corrections just to be clear-- the force is not proportional to distance, but rather 1/d2, and Einstein's treatment of gravity does not have it being a force at all-- he says that this weird coincidence that the force scales the same way as the inertia is telling us that it isn't a force at all, it is an accelerating coordinate system that automatically makes all objects seen from it appear to accelerate the same. The reason we don't notice that the coordinate system is accelerating is that we are imagining a flat spacetime, but that's the wrong geometry-- gravity curves the spacetime, and not realizing this causes us to choose accelerated coordinates without knowing it. In other words, the acceleration of a falling object is an illusion-- we are really the ones who are accelerating, not the falling object.)
    Last edited: Sep 14, 2011
  5. Sep 14, 2011 #4
    So the free-falling rate of an object toward the earth is solely dependent on the earth's gravitational field and not it's own?
    Last edited: Sep 14, 2011
  6. Sep 14, 2011 #5

    Ken G

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    That brings in rcgldr's point-- we are always making idealizations in physics. When we consider a falling object, we usually treat it as a "test particle", i.e., a particle that does not change its environment but is acted upon by its environment. Basically that requires the mass of the object be a whole lot less than the mass of Earth, and then the answer to your question is "yes." But, if the mass is not that small, or we are doing incredibly precise calculations, then in Newtonian physics we'd have to account for the Earth's acceleration upward due to the gravity of the object, and the free-fall rate relative to Earth must include the Earth's motion. In Einstein's model of gravity, the situation is even more complicated, because both objects contribute to the curvature of spacetime, and the effects are nonlinear-- i.e., you can't just add the two together to get the result. But to worry about that for most falling objects, you would need to be interested in super high accuracy, I couldn't tell if you wanted that level of precision.
  7. Sep 14, 2011 #6
    F being the gravitational force of the objects gravitational field I suppose? Does mass really resist the free-fall of another object's gravitational field? Would Newton's law apply to general relativity where objects are really just free-falling at different rates depending on their mass?
  8. Sep 14, 2011 #7


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    Mass doesn't resist free fall, it resists acceleration due to forces. If the only force acting is proportional to mass, all objects accelerate the same way.

    They are not doing that in general relativity. They all fall at the same rate.
  9. Sep 14, 2011 #8
    In general relativity (closest theory to reality) gravity is not a "force" so I'd like to keep Newton's approximations aside. If the speed of free-fall of an object toward a massive body is due to the mass of both bodies as both contribute to the curvature of space-time, and mass does not resist free-fall, then can't we say that two different masses free-fall at two different rates in general relativity?

    I actually meant to say "objects cause objects to free-fall at different rates depending on their mass". So to be clear, and I'm not very interested in idealizations, the rate of free-fall of an object toward the earth is solely dependent on the earths gravitational field and it's own only pulls earth toward itself? Can this just be seen as the object free-falling at rates depending on it's mass anyway?
  10. Sep 14, 2011 #9


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    Gravity is still a force in GR, but unlike in Newton's theory it is an inertial force, not an interaction force. And just like all inertial forces it accelerates every object in the same way regardless of the objects mass.

    Not sure what that means

    The coordinate accelerations depend on the reference frame. In the inertial rest frame of the common center of mass, the acceleration of each body depends only on the mass of the other body.
  11. Sep 14, 2011 #10
    I'm just asking if due to the space-time warping contribution of an object falling toward earth, even if virtual to us, can the object be seen to be accelerating in the reference frame of an object falling next to it with less mass and less warping contribution?
    Last edited: Sep 14, 2011
  12. Sep 14, 2011 #11


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    Please start by reading the FAQ subforum in the General Physics forum.

    https://www.physicsforums.com/forumdisplay.php?f=209 [Broken]

    Last edited by a moderator: May 5, 2017
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